4
$\begingroup$

Assuming $A$ and $B$ are two non-negative real-valued random variables such that

  1. $\mathrm{E}(A)=\mathrm{E}(B)$ (equal means)
  2. $\mathrm{Var}(A)=\mathrm{Var}(B)<\epsilon$ (equal small variances)

is there a way to prove that $\frac{1}{N}\sum_{j=1}^Ne^{-a_{j}}$ and $\frac{1}{N}\sum_{j=1}^Ne^{-b_{j}}$ are arbitrarily close to each other where $a_j$ and $b_j$ are realizations taken from $A$ and $B$, respectively. ($N$ can be assumed to be large as well)

$\endgroup$
6
$\begingroup$

Although it is not explicitly specified, I will assume that you intend for all the realisations of these random variables to be independent (i.e., I will assume joint independence of all the random variables in both series). The difference between the two series is the random variable defined by the function:

$$D(N) = \frac{1}{N} \sum_{i=1}^N (e^{-A_i} - e^{-B_i}).$$

Since $e^{-a} \leqslant 1$ for all $a \geqslant 0$, it follows that $\mathbb{V}(e^{-A}) \leqslant \mathbb{E}(e^{-2A}) \leqslant 1$ for any non-negative random variable $A$. Thus, we have:

$$\begin{equation} \begin{aligned} \mathbb{V}(D(N)) &= \frac{1}{N^2} \sum_{i=1}^N \Big( \mathbb{V}(e^{-A_i}) + \mathbb{V}(e^{-B_i}) \Big) \\[6pt] &\leqslant \frac{1}{N^2} \sum_{i=1}^N \Big( 1 + 1 \Big) \\[6pt] &= \frac{1}{N^2} \cdot 2N \\[6pt] &= \frac{2}{N}. \\[6pt] \end{aligned} \end{equation}$$

We therefore have $\lim_{N \rightarrow \infty} \mathbb{V}(D(N)) = 0$, so the variance of the difference converges to zero. We also have the constant mean difference:

$$\begin{equation} \begin{aligned} \mathbb{E}(D(N)) &= \frac{1}{N} \sum_{i=1}^N \Big( \mathbb{E}(e^{-A_i}) - \mathbb{E}(e^{-B_i}) \Big) \\[6pt] &= \mathbb{E}(e^{-A}) - \mathbb{E}(e^{-B}). \\[6pt] \end{aligned} \end{equation}$$

Combining these results we see that $D(N)$ converges in mean square to $\mathbb{E}(e^{-A}) - \mathbb{E}(e^{-B})$, which is a constant value. Since the variances of both random variables are small, this limiting value should be close to (but not necessarily equal to) zero. Thus, for large values of $N$ you would indeed expect the difference to converge towards a value near zero.

$\endgroup$
  • $\begingroup$ Thank you for the answer. Is there a mathematical way to show that E(exp(-A))-E(exp(-B)) is close to zero in the end? Also, have you used anywhere the fact that means and variances of A and B are the same? $\endgroup$ – nOp May 1 '19 at 4:58
  • $\begingroup$ Under the conditions you have specified, the variance of both random variables are low, so they are distributed steeply around their mean ---i.e., you have $A \approx B \approx \mu$. This gives $\mathbb{E}(e^{-A}) \approx \mathbb{E}(e^{-B})$, which means that the difference should be near zero. $\endgroup$ – Reinstate Monica May 1 '19 at 9:01
  • $\begingroup$ I got back to this after a while. When you are computing the difference D(N), A_i's are realizations and therefore constant, right? Then, what does it mean when you write variance of exp(A_i), i.e., V(exp(A_i))? As variance of a constant value is zero, i can not make sense out of D(N). It would be great if you could clarify this. $\endgroup$ – nOp Jun 14 '19 at 23:22
  • $\begingroup$ For the record, note that the @ben 's answer above is NOT an answer to the question. $\endgroup$ – nOp Jun 17 '19 at 18:42
2
$\begingroup$

Just apply the law of large numbers, since $e^{-a_j}$ has finite variance because $a$ is positive and has finite variance it self. The result is that the variance of $e^{-a_j}$ is smaller than the variance of $a$. This is elaborated in this question: https://mathoverflow.net/questions/330348/proof-of-variance-bounds-for-transformed-random-variables/330357#330357

$\endgroup$
  • $\begingroup$ Thanks. LLN says the sums equal E(exp(-A)) and E(exp(-B)), respectively. What can then be said about these two means? $\endgroup$ – nOp Apr 30 '19 at 20:09
  • $\begingroup$ $\frac{1}{N}\sum{exp(-a_i)}->E(exp(-A))$, when N goes to infinity $\endgroup$ – Peter Mølgaard Pallesen May 1 '19 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.